Can you store a googolplex digitally?

[lovenugget]

will someone please enlighten me on whether this is possible? I'm debating a friend on whether it is, but I'm starting to reconsider my position. to be more specific, i want to know if you can store (10^(10^100)) in a text file on a hard drive that would fit inside the universe. also, what's that in bytes (lol). thanks.

 

[Muphrid]

We estimate the number of atoms in the known universe to be about $10^{80}$, which is far short of a googol, let alone a googolplex. So even if you could use every atom in the universe as a bit, you wouldn't get close to all the digits necessary.

 

[russ_watters]

Do you mean just the number or actually that many pieces of data? Because the number already appears in this thread...

 

[jtbell]

When I saw the thread title I first thought you were asking about Google's headquarters.

http://www.time.com/time/photogallery/0,29307,1947844,00.html

 

[russ_watters]

Heh, me too.

 

[lovenugget]

you're right. good answer. what if i asked instead if you could record all of the zeros of a googleplex? i understand the number of zeros is far less.

 

[russ_watters]

You cannot write the number out, but since it is a simple number, the sci notation is just as good. I'd declare yourself the winner. I'll pm you the address to mail my commission check.

 

[mfb]

you're right. good answer. what if i asked instead if you could record all of the zeros of a googleplex? i understand the number of zeros is far less.

In base 10, this would require a 10^100 bytes of storage: No, you cannot, at least not in the classical way. You can use compression, of course - and there are compression systems which can do this with a conventional hard disk.

In an arbitrary base: In base googoloplex, your number is 10. Easy to store in conventional ways.

Googleplex is something different, as you can see from the images ;).

 

[lovenugget]

oh sorry about the error in the title. it's correctly spelled as you all have. thanks for the responses. i still wonder though, is there a googolplex photons in the universe? there has to be... or else I'm still underestimating the true size of the how large a number it is.

EDIT

there's not.

*head explodes*.

 

[Kholdstare]

actually russ the precision has to be sacrificed when someone tries to save such big number. IEEE 754 floating point standard is good example here. In fact you can save a very large number with a bad precision for a minimum number of bits. The more you add bits into it the precision improves (and also you can use the same memory to store even larger number with even worse precision). Talking about trade-off, ha.

http://en.wikipedia.org/wiki/IEEE_754

 

[Dembadon]

In base 10, this would require a 10^100 bytes of storage: [...].

Time for a new integer data type! Tinyints just aren't cuttin' the mustard anymore.

 

[mfb]

i still wonder though, is there a googolplex photons in the universe?

The whole universe? It might be infinite, in this case: Sure
The observable universe? Not even close. With ~10⁸⁰ atoms, the number of photons is larger by many orders of magnitude. However, "many orders of magnitude" change that to ~10⁹⁰, ~10¹⁰⁰ or maybe even ~10¹⁵⁰ photons.
You think that there are 10¹⁵⁰ photons for every photon in my upper estimate? Then we would have 10³⁰⁰ photons.
As you can see, the exponent grows slowly even with extremely high estimates - it will not reach 100...0 (100 zeros).

 

[russ_watters]

actually russ the precision has to be sacrificed when someone tries to save such big number. IEEE 754 floating point standard is good example here. In fact you can save a very large number with a bad precision for a minimum number of bits. The more you add bits into it the precision improves (and also you can use the same memory to store even larger number with even worse precision). Talking about trade-off, ha.

http://en.wikipedia.org/wiki/IEEE_754

What I was getting at when I said it is a simple number is that with only 1 sig fig, there is no precision to be sacrificed by storing it in some variation of sci notation.

 

[-Job-]

It may be possible to store such a large value within the Universe, though it's likely to be at least a computationally difficult problem and we would possibly have to give up the ability to store any given string of 10¹⁰¹⁰⁰.

Given any two particles, there is a value implicitly stored that corresponds to their distance, in three dimensions, measured in some unit. If we measure at the Planck length, that's approximately 100 (?) bits per distance.

Given n particles, you can potentially store 2ⁿ such distances, amounting to a maximum of 100 * 2ⁿ bits, which would be hard or impossible to reach depending on the value being stored, and quite an optimization problem.

With 2²⁶⁵ particles in the universe, this would yield at best 100 * 2²²⁶⁵ - huge, but not as large as 10¹⁰¹⁰⁰.

 

[mfb]

You cannot get exponential memory with n particles, those 2ⁿ distances are not independent. Neglecting quantum mechanics, you can fully describe all via their coordinates - 3 per particle in space, 6 if you include their momentum. With ~100 bits per coordinate (roughly 1m^3 of space), 10^30 particles would give ~10^34 bits of storage. If you use the whole accessible universe, you get 200 bits per coordinate and ~10^80 particles, for a total capacity of ~10^84 bits.
Quantum mechanics does not allow that precision, so it gets even worse.

 

[-Job-]

You cannot get exponential memory with n particles, those 2ⁿ distances are not independent. Neglecting quantum mechanics, you can fully describe all via their coordinates - 3 per particle in space, 6 if you include their momentum. With ~100 bits per coordinate (roughly 1m^3 of space), 10^30 particles would give ~10^34 bits of storage. If you use the whole accessible universe, you get 200 bits per coordinate and ~10^80 particles, for a total capacity of ~10^84 bits.
Quantum mechanics does not allow that precision, so it gets even worse.

That's right, as I pointed out only some strings may be encoded to make maximum use of the 2ⁿ values and it becomes a large optimization problem.

The point is there's substantially more storage capacity in the universe than the number of particles. Given two strings, we can encode a number of bits corresponding to the length of the universe measured in Planck units - I haven' t computed this value but it's greater than 100 bits.

 

[mfb]

2¹⁰⁰ Planck lengths are some micrometers, 2²⁰⁰ Planck lengths are similar to the size of the accessible universe.

The point is there's substantially more storage capacity in the universe than the number of particles. Given two strings, we can encode a number of bits corresponding to the length of the universe measured in Planck units - I haven' t computed this value but it's greater than 100 bits.

It is not substantially. While 10^80 and 10^84 differ by a factor of 10000, this is a small factor compared to the overall size of the numbers.

 

[-Job-]

Here's an example - please check my math, I'm on my phone.

Given 2^265 (approx 10^80) particles in the universe, suppose we split the particles into groups of three. Each three particles store three distances - these values are independent, given that there are 3 dimensions. Let each distance encode 100 bits, as a low estimate.

Each three particles has 2^300 possible states. Let each group of three particles correspond to a digit, in base 2^300 encoding, a really long alphabet. We can store (2^265)/3 such digits - approx 2^260.

The result would be the enormous value of (2^300)^(2^260).

 

[mfb]

They are not independent, they have to satisfy the triangle inequality and 3 particles are always within a plane. But well, that does not matter.

Your result is the number of different states you can encode with this (quite inefficient) way to store data. It can be simplified to $2^{300*2^{260}} \approx 2^{2^{268}}$. This corresponds to $2^{268} \approx 5\cdot 10^{80}$ bits of storage. Less than a direct position-dependent storage (which needs 4 particles to define a coordinate system, but who cares about 4 out of 10⁸⁰ particles).

 

[akdude1]

Theoretically its possible if you had enough things to store information on. Except did this discussion turn into whether or not there are even enough particles in the Universe to store that information.

 

[Greg Bernhardt]

Changed the title to "Googolplex" :)

 

[IsometricPion]

The arxiv paper linked here indicates that the maximum amount of data that can be processed is no more than of order 10^120 (another paper linked here indicates that most of this data is encoded in gravitational degrees of freedom). So it seems that in principle there may be ways to store the of order 10^100 bits necessary to write 10^(10^100) in binary. In fact, it should be possible with a few hundred billion solar masses. That being said, I'm not sure that the best use of a medium-large sized galaxy worth of mass-energy is to collapse (very carefully) into a black hole like state in order to store a single number.

 

[.Scott]

According to Jacob Bekenstein:
The visible universe contains at least 10^100 bits of entropy, which could in principle be packed inside a sphere a tenth of a light-year across. Estimating the entropy of the universe is a difficult problem, however, and much larger numbers, requiring a sphere almost as big as the universe itself, are entirely plausible.

So, yes!

 

[jaxxrr]

meanwhile some entity is havin' a laugh seeing as they store our universe simulation on some lame computer.. with 10^1000 storage capacity or something..